A battery of emf `epsi` and internal resistance r, when connected across an external resistance of 12 `Omega` , produces a current of 0.5 A. When connected across a resistance of 25`Omega`, it produces a current of 0.25 A. Determine the emf and internal resistance of the cell.

Explanation

We know that current drawn `I = (epsi)/(R + r)`
As per question when external resistance `R_1 = 12Omega` current `I_1`=0.5 A, and for resistance `R_2 = 25Omega`, current `I_2 = 0.25 A` .
` therefore 0.5 = (epsi)/(12 + r) ` ....(i) and `0.25 = (epsi)/(25+ r)` .....(ii)
From (i) and (ii) , we get `r = 1 Omega ` and `epsi = 6.5 V`